3.53 \(\int \frac{\log (c (a+b \sqrt{x})^p)}{x^4} \, dx\)

Optimal. Leaf size=130 \[ -\frac{b^3 p}{9 a^3 x^{3/2}}+\frac{b^2 p}{12 a^2 x^2}-\frac{b^5 p}{3 a^5 \sqrt{x}}+\frac{b^4 p}{6 a^4 x}+\frac{b^6 p \log \left (a+b \sqrt{x}\right )}{3 a^6}-\frac{b^6 p \log (x)}{6 a^6}-\frac{\log \left (c \left (a+b \sqrt{x}\right )^p\right )}{3 x^3}-\frac{b p}{15 a x^{5/2}} \]

[Out]

-(b*p)/(15*a*x^(5/2)) + (b^2*p)/(12*a^2*x^2) - (b^3*p)/(9*a^3*x^(3/2)) + (b^4*p)/(6*a^4*x) - (b^5*p)/(3*a^5*Sq
rt[x]) + (b^6*p*Log[a + b*Sqrt[x]])/(3*a^6) - Log[c*(a + b*Sqrt[x])^p]/(3*x^3) - (b^6*p*Log[x])/(6*a^6)

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Rubi [A]  time = 0.0794123, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2454, 2395, 44} \[ -\frac{b^3 p}{9 a^3 x^{3/2}}+\frac{b^2 p}{12 a^2 x^2}-\frac{b^5 p}{3 a^5 \sqrt{x}}+\frac{b^4 p}{6 a^4 x}+\frac{b^6 p \log \left (a+b \sqrt{x}\right )}{3 a^6}-\frac{b^6 p \log (x)}{6 a^6}-\frac{\log \left (c \left (a+b \sqrt{x}\right )^p\right )}{3 x^3}-\frac{b p}{15 a x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*Sqrt[x])^p]/x^4,x]

[Out]

-(b*p)/(15*a*x^(5/2)) + (b^2*p)/(12*a^2*x^2) - (b^3*p)/(9*a^3*x^(3/2)) + (b^4*p)/(6*a^4*x) - (b^5*p)/(3*a^5*Sq
rt[x]) + (b^6*p*Log[a + b*Sqrt[x]])/(3*a^6) - Log[c*(a + b*Sqrt[x])^p]/(3*x^3) - (b^6*p*Log[x])/(6*a^6)

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\log \left (c \left (a+b \sqrt{x}\right )^p\right )}{x^4} \, dx &=2 \operatorname{Subst}\left (\int \frac{\log \left (c (a+b x)^p\right )}{x^7} \, dx,x,\sqrt{x}\right )\\ &=-\frac{\log \left (c \left (a+b \sqrt{x}\right )^p\right )}{3 x^3}+\frac{1}{3} (b p) \operatorname{Subst}\left (\int \frac{1}{x^6 (a+b x)} \, dx,x,\sqrt{x}\right )\\ &=-\frac{\log \left (c \left (a+b \sqrt{x}\right )^p\right )}{3 x^3}+\frac{1}{3} (b p) \operatorname{Subst}\left (\int \left (\frac{1}{a x^6}-\frac{b}{a^2 x^5}+\frac{b^2}{a^3 x^4}-\frac{b^3}{a^4 x^3}+\frac{b^4}{a^5 x^2}-\frac{b^5}{a^6 x}+\frac{b^6}{a^6 (a+b x)}\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{b p}{15 a x^{5/2}}+\frac{b^2 p}{12 a^2 x^2}-\frac{b^3 p}{9 a^3 x^{3/2}}+\frac{b^4 p}{6 a^4 x}-\frac{b^5 p}{3 a^5 \sqrt{x}}+\frac{b^6 p \log \left (a+b \sqrt{x}\right )}{3 a^6}-\frac{\log \left (c \left (a+b \sqrt{x}\right )^p\right )}{3 x^3}-\frac{b^6 p \log (x)}{6 a^6}\\ \end{align*}

Mathematica [A]  time = 0.0627276, size = 114, normalized size = 0.88 \[ \frac{a b p \sqrt{x} \left (-20 a^2 b^2 x+15 a^3 b \sqrt{x}-12 a^4+30 a b^3 x^{3/2}-60 b^4 x^2\right )-60 a^6 \log \left (c \left (a+b \sqrt{x}\right )^p\right )+60 b^6 p x^3 \log \left (a+b \sqrt{x}\right )-30 b^6 p x^3 \log (x)}{180 a^6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*Sqrt[x])^p]/x^4,x]

[Out]

(a*b*p*Sqrt[x]*(-12*a^4 + 15*a^3*b*Sqrt[x] - 20*a^2*b^2*x + 30*a*b^3*x^(3/2) - 60*b^4*x^2) + 60*b^6*p*x^3*Log[
a + b*Sqrt[x]] - 60*a^6*Log[c*(a + b*Sqrt[x])^p] - 30*b^6*p*x^3*Log[x])/(180*a^6*x^3)

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Maple [F]  time = 0.067, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}}\ln \left ( c \left ( a+b\sqrt{x} \right ) ^{p} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(a+b*x^(1/2))^p)/x^4,x)

[Out]

int(ln(c*(a+b*x^(1/2))^p)/x^4,x)

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Maxima [A]  time = 1.12465, size = 132, normalized size = 1.02 \begin{align*} \frac{1}{180} \, b p{\left (\frac{60 \, b^{5} \log \left (b \sqrt{x} + a\right )}{a^{6}} - \frac{30 \, b^{5} \log \left (x\right )}{a^{6}} - \frac{60 \, b^{4} x^{2} - 30 \, a b^{3} x^{\frac{3}{2}} + 20 \, a^{2} b^{2} x - 15 \, a^{3} b \sqrt{x} + 12 \, a^{4}}{a^{5} x^{\frac{5}{2}}}\right )} - \frac{\log \left ({\left (b \sqrt{x} + a\right )}^{p} c\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b*x^(1/2))^p)/x^4,x, algorithm="maxima")

[Out]

1/180*b*p*(60*b^5*log(b*sqrt(x) + a)/a^6 - 30*b^5*log(x)/a^6 - (60*b^4*x^2 - 30*a*b^3*x^(3/2) + 20*a^2*b^2*x -
15*a^3*b*sqrt(x) + 12*a^4)/(a^5*x^(5/2))) - 1/3*log((b*sqrt(x) + a)^p*c)/x^3

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Fricas [A]  time = 2.27093, size = 269, normalized size = 2.07 \begin{align*} -\frac{60 \, b^{6} p x^{3} \log \left (\sqrt{x}\right ) - 30 \, a^{2} b^{4} p x^{2} - 15 \, a^{4} b^{2} p x + 60 \, a^{6} \log \left (c\right ) - 60 \,{\left (b^{6} p x^{3} - a^{6} p\right )} \log \left (b \sqrt{x} + a\right ) + 4 \,{\left (15 \, a b^{5} p x^{2} + 5 \, a^{3} b^{3} p x + 3 \, a^{5} b p\right )} \sqrt{x}}{180 \, a^{6} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b*x^(1/2))^p)/x^4,x, algorithm="fricas")

[Out]

-1/180*(60*b^6*p*x^3*log(sqrt(x)) - 30*a^2*b^4*p*x^2 - 15*a^4*b^2*p*x + 60*a^6*log(c) - 60*(b^6*p*x^3 - a^6*p)
*log(b*sqrt(x) + a) + 4*(15*a*b^5*p*x^2 + 5*a^3*b^3*p*x + 3*a^5*b*p)*sqrt(x))/(a^6*x^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(a+b*x**(1/2))**p)/x**4,x)

[Out]

Timed out

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Giac [B]  time = 1.32791, size = 437, normalized size = 3.36 \begin{align*} -\frac{\frac{60 \, b^{7} p \log \left (b \sqrt{x} + a\right )}{{\left (b \sqrt{x} + a\right )}^{6} - 6 \,{\left (b \sqrt{x} + a\right )}^{5} a + 15 \,{\left (b \sqrt{x} + a\right )}^{4} a^{2} - 20 \,{\left (b \sqrt{x} + a\right )}^{3} a^{3} + 15 \,{\left (b \sqrt{x} + a\right )}^{2} a^{4} - 6 \,{\left (b \sqrt{x} + a\right )} a^{5} + a^{6}} - \frac{60 \, b^{7} p \log \left (b \sqrt{x} + a\right )}{a^{6}} + \frac{60 \, b^{7} p \log \left (b \sqrt{x}\right )}{a^{6}} + \frac{60 \,{\left (b \sqrt{x} + a\right )}^{5} b^{7} p - 330 \,{\left (b \sqrt{x} + a\right )}^{4} a b^{7} p + 740 \,{\left (b \sqrt{x} + a\right )}^{3} a^{2} b^{7} p - 855 \,{\left (b \sqrt{x} + a\right )}^{2} a^{3} b^{7} p + 522 \,{\left (b \sqrt{x} + a\right )} a^{4} b^{7} p - 137 \, a^{5} b^{7} p + 60 \, a^{5} b^{7} \log \left (c\right )}{{\left (b \sqrt{x} + a\right )}^{6} a^{5} - 6 \,{\left (b \sqrt{x} + a\right )}^{5} a^{6} + 15 \,{\left (b \sqrt{x} + a\right )}^{4} a^{7} - 20 \,{\left (b \sqrt{x} + a\right )}^{3} a^{8} + 15 \,{\left (b \sqrt{x} + a\right )}^{2} a^{9} - 6 \,{\left (b \sqrt{x} + a\right )} a^{10} + a^{11}}}{180 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b*x^(1/2))^p)/x^4,x, algorithm="giac")

[Out]

-1/180*(60*b^7*p*log(b*sqrt(x) + a)/((b*sqrt(x) + a)^6 - 6*(b*sqrt(x) + a)^5*a + 15*(b*sqrt(x) + a)^4*a^2 - 20
*(b*sqrt(x) + a)^3*a^3 + 15*(b*sqrt(x) + a)^2*a^4 - 6*(b*sqrt(x) + a)*a^5 + a^6) - 60*b^7*p*log(b*sqrt(x) + a)
/a^6 + 60*b^7*p*log(b*sqrt(x))/a^6 + (60*(b*sqrt(x) + a)^5*b^7*p - 330*(b*sqrt(x) + a)^4*a*b^7*p + 740*(b*sqrt
(x) + a)^3*a^2*b^7*p - 855*(b*sqrt(x) + a)^2*a^3*b^7*p + 522*(b*sqrt(x) + a)*a^4*b^7*p - 137*a^5*b^7*p + 60*a^
5*b^7*log(c))/((b*sqrt(x) + a)^6*a^5 - 6*(b*sqrt(x) + a)^5*a^6 + 15*(b*sqrt(x) + a)^4*a^7 - 20*(b*sqrt(x) + a)
^3*a^8 + 15*(b*sqrt(x) + a)^2*a^9 - 6*(b*sqrt(x) + a)*a^10 + a^11))/b